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I am trying to compute the group structure as the title says of $E:y^2=x^3 - x$ over $\mathbb{F}_p$ with $p\equiv 1\bmod 8$ and $p-1$ a square, that is, $p=(a+i)(a-i)=a^2+1$ over $\mathbb{Z}[i]$.

I just proved that $\#E(\mathbb{F}_p)=p + 1 - 2 = p - 1 = a^2$ and in MAGMA it says that as an abstract group for all my examples is isomorphic to $\mathbb{Z}/(a)\times \mathbb{Z}/(a)$. I spent some time but I cannot get more far than the cardinality yet (I thought it was going to be simpler or there is something I am not seeing).

This curve in this case has CM by $i$ that is, $\text{End}_{\mathbb{F}_p}(E)=\mathbb{Z}[i]$. Another trivial observation is that it has full $2$-torsion, in fact its Mordell-Weil group is isomorphic to $\mathbb{Z}/(2)\times\mathbb{Z}/(2)$ so this group should be a subgroup of $E(\mathbb{F}_p)$ for all $p>2$, hence $E(\mathbb{F}_p)$ is not cyclic.

I am using $p\equiv 1\bmod 8$ just to fix the cardinality to be always $\#E(\mathbb{F}_p)=a^2$.

Any hints will be appreciated, maybe I have to use the Weil Pairing, since it could be easier to prove that $X^a - 1$ splits over $\mathbb{F}_p[X]$ and use the Weil pairing is onto $\mu_a$. But also I am a little stucked here.

  • $\begingroup$ If $p=a^2+1$ then $X^a-1$ divides $X^p-X=(X^{a^2}-1)X$ whose roots are exactly the members of $\mathbb{F}_p$ by Fermat's little theorem, so $X^a-1$ splits, which should be enough to give you the Weil pairing $\endgroup$ – W. Cadegan-Schlieper Jul 20 '17 at 0:16
  • 3
    $\begingroup$ I think it follows from your calculation that the trace of Frobenius is $2$ and Frobenius divides $p$ that Frobenius must be $ai + 1$ upon identifying the endomorphism ring with $\mathbb{Z}[i]$. It follows that the fixed points of Frobenius are killed by $a$. $\endgroup$ – Felipe Voloch Jul 20 '17 at 7:31

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